Thank you for your attention.
My problem is
$U_x+\frac{Uy}{U} = U^2,\;\; U(x,1) = 1$.
I will really appreciate any clue in solving problems like this.
Am I right with :
$\frac{dx}{1}$ = $udy$ = $\frac{du}{u^2}$
if yes, then I am confused with how to solve these three curves intersection.
$$u_x+\frac{u_y}{u} = u^2$$
$\frac{dx}{1}=udy=\frac{du}{u^2}\quad$ is OK.
First characteristic equation, from $\frac{dx}{1}=\frac{du}{u^2}$ : $$\frac{1}{u}+x=c_1$$
Second characteristic equation, from $udy=\frac{du}{u^2}$ : $$\frac{1}{2u^2}+y=c_2$$
General solution : $$\frac{1}{2u^2}+y=F\left(\frac{1}{u}+x\right)$$ $F$ is an arbitrary function to be determined according to boundary condition.
Boundary condition : $\quad u(x,1)=1$ .
This is a special case because the boundary condition is defined on a characteristic curve : $\quad \frac{1}{2u^2}+y=\frac{1}{2}+1=\frac{3}{2}=c_2$ .
$\frac{1}{u}+x=1+x$ and the solution $\frac{1}{2u^2}+y=F\left(\frac{1}{u}+x\right)$ is only possible if the function $F$ is a constant function : $\frac{1}{2u^2}+y=\frac32$
Finally, the solution is not function of $x$ : $$u(x,y)=\frac{1}{\sqrt{3-2y}}$$ To check, put it into the PDE.